Evaluate the integral without using tables.

$\int}_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}$

BenoguigoliB
2021-11-08
Answered

Evaluate the integral without using tables.

$\int}_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}$

You can still ask an expert for help

yagombyeR

Answered 2021-11-09
Author has **92** answers

Step 1

Given,

$I={\int}_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}}$

Step 2

Formula used:

$\int \frac{dx}{\sqrt{1-{x}^{2}}}={\mathrm{sin}}^{-1}\left(x\right)+C$

$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$

Step 3

Apply the above formula, we get

$I={\int}_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}}$

$={\left[{\mathrm{sin}}^{-1}\left(x\right)\right]}_{0}^{1}$

$={\mathrm{sin}}^{-1}\left(1\right)-{\mathrm{sin}}^{-1}\left(0\right)$

$={\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{\pi}{2}\right)\right)-{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(0\right)\right)$

$=\frac{\pi}{2}-0$

$=\frac{\pi}{2}$

Given,

Step 2

Formula used:

Step 3

Apply the above formula, we get

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